Question

If $A$ and $B$ are two events such that
$P(A / B)=0.6, P(B / A)=0.3$ and $P(A)=0.1$
then $P(\bar{A} \cap \bar B)$ equals

• A. 0.88
• B. 0.12
• C. 0.6
• D. 0.4

Answer 1

Answer: (A)

We have,
$P(A / B)=0.6, P(B / A)=0.3, P(A)=0.1$
We know that,
$\Rightarrow P(B / A) =\frac{P(B \cap A)}{P(A)} $
$0.3 =\frac{P(B \cap A)}{0.1} $
$\Rightarrow P(B \cap A)=0.03$
Or $P(A \cap B)=0.03$
Also,$ P(A / B) =\frac{P(A \cap B)}{P(B)} $
$ \Rightarrow 0.6 =\frac{0.03}{P(B)}$
$\Rightarrow P(B)=0.05$
$\therefore P(\bar{A} \cap \bar{B})=1-P(A \cup B)$
$=1-[P(A)+P(B)-P(A \cap B)]$
$=1-0.1-0.05+0.03=0.88$